## Linear Algebraic Groups by Armand Borel PDF

By Armand Borel

ISBN-10: 1461269547

ISBN-13: 9781461269540

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**Example text**

We then say that CI. ) if, for each such V', the induced morphism V' -> W' (which is dominant) has the corresponding property. ( V') = W' is an irreducible subvariety of W, and it will be an irreducible component of W provided it contains a non-empty open set in W. Since V' contains such an open set in V this remark shows that: /1' 'J. is sUijecl it:e and open I hen "Y. is dominant. As a converse, if Wand V are irreducible (for convenience), given an injective homomorphism /3: K(W)-> K(V), there is a dominant morphism a of a Zariski open subset U of V into W, such that /3 = aU.

QA/A 'dAJ, M). Q A modulo the A-module generated by all dfU EJ). ,s over a set of generators of 1. For example, suppose A = k[Tl , ... , Tn] is a polynomial ring, so that A'=k[tl, ... ,tn ] (tj=image of TJ Then iffl, ... Q A' is defined by generators dt j (I ~ i ~ n) and relations I(OJ1)U)dt aT i j i =O (I~j~m). 4 Proposition. e. that k is mapped onto A' = AjJ. QA- Proof. e. mod(J/J 2,M). If X:A ~ M is a k-derivation then X(k) = 0 so, since A = k(£)J, X is determined by XIJ. Since J M = 0 we must have X(J2) = 0, so X is determined by a homomorphism h:J /12 ~ M.

B] (a): V, §16, no. 6, Cor. I). B] (a): III, §1O, no. 7) 0--+ DerA(B, B)--+ Derk(B, B)--+ Derk(A, B), by counting B-dimensions, that B is separable over A=- Derk(B, B)--+ Derk(A, B) is surjective =-B@QA--+QBis injective. 7 Tensor products. Suppose A = A]@A 2, and write Q j = Q A;. Then k Equivalently, if M is any A-module, we have Derk(A, M) ~ Derk(A], M)EBDer k(A 2 , M). The map from left to right is induced by the homomorphisms A j --+ A. For the inverse we must produce a k-derivation X: A --+ M from a given pair of them Xj:Aj--+M.

### Linear Algebraic Groups by Armand Borel

by William

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